1. Magnetic field of a steady current Section 30

2. Previously supposed zero net current. Then Now let the net current j be non-zero. Then “Conduction” current density No contribution to net current Rate of energy dissipation per unit volume = E. j

3. Distribution of j is determined by The H-field generated by j is not included if From section 21 Hence, H is determined from the given j. Recipe: Find j. Then find H. Ignore the effect that H has on j.

4. H determined from given j.

5. Boundary condition H 1t = H 2t still holds Proof: Homogeneous interface between conductors. For homogeneous interface, Is finite J is finite everywhere, including boundaries finite H y is continuous. (H t is continuous.)

6. For the given j, we want to find H and B These are the equations we need to solve. Introduce the vector potential and find the equation for it. Due to gauge invariance, we can place a condition on A, namely Assume isotropic linear medium, or else there are too many unknowns A differential equation for A with j as the source. Recipe: Solve for A, then find B, then H.

7. Special case: homogeneous medium where  is spatially uniform Differential equation for A with given j source term Recipe: Find A, then find B, then H.

8. Special case: Piecewise homogeneous medium Different parts have different  holds in each part. Boundary condition between the parts is continuous.

9. Special case.: 2D problems, i.e. systems with tranlational invariance along one axis, like a straight wire. z y By assumption One possibility for current: Then we expect: holds automatically Condition Magnetic field likes in the xy plane

10. What equation does A satisfy in this 2D problem? Homework But Equation for scalar A when  =  (x,y) and j = j(xy)e z

11. What if 2D problem is piecewise homogeneous? holds in each region with own  Boundary conditions are continuity of A,, and (curlA) t /  Homework

12. Special case of 2D problem with current along axis: cylindrical symmetry. If current is symmetrical about the z-axis, j z = j(r) Then magnetic field lines are circles (r = constant) (the field must have the same symmetry as its sources. r = const Any surface bound by contour r = const Stokes Total current enclosed within radius r

13. Solve for A, then Another kind of 2D problem is the axially symmetric current distribution We expect Magnetic induction (field) H-field

14. Non magnetic media:  = 1 Now there are no conditions at interfaces or surfaces since all media are magnetically identical (v.2 section 43) Since  = 1, H = B Acts on coordinates r of field point P HW

15. Thin wires, neglect internal regions of wire: “linear currents” Total current in the conductor R3R3 Biot-Savart law.

16. If thin conducting material is surrounded by magnetic material of permeability 

17. If j = 0 except for linear currents Except for line singularities Then Some magnetic scalar potential Contrast with electrostatics grad  has same value at start and end of integral Same holds for  :  is single valued. In all space